A129505 Number of permutations of 2n-1 objects with exactly n cycles.

1, 3, 35, 735, 22449, 902055, 44990231, 2681453775, 185953177553, 14710753408923, 1307535010540395, 129006659818331295, 13990945200239106865, 1654339178844590073615, 211821088794711294496815, 29197210605623737977801375, 4310704065427058593776844065

Offset: 1

Author

Paul D. Hanna, Apr 18 2007

Keywords

• nonn

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..200
• D. Kruchinin and V. Kruchinin, A Method for Obtaining Generating Function for Central Coefficients of Triangles, Journal of Integer Sequence, Vol. 15 (2012), article 12.9.3.

Crossrefs

Cf. A008275, A129506.

Cf. A238685.

Programs

Haskell

a129505 n = abs $ a008275 (2 * n - 1) n -- Reinhard Zumkeller, Mar 02 2014

Mathematica

t[n_] := SymmetricPolynomial[n, Range[1, 2 n]]
Table[t[n], {n, 1, 6}]  (* A129505 *)
(* Clark Kimberling, Dec 30 2011 *)
Table[Abs[StirlingS1[2*n-1, n]], {n, 1, 20}] (* Vaclav Kotesovec, Dec 28 2013 *)

Maxima

a(n):=((2*n+1)*(-1)^n*((sum((stirling1(2*i-1,i)*binomial(2*n,2*i-1)* stirling1(2*(n-i)+1,n-i))/((n-i)*binomial(n,i)),i,1,n-1)) -n*stirling1(2*n-1,n) +stirling1(2*n,n)))/(n+1); /* Vladimir Kruchinin, Feb 28 2013 */

Maxima

a(n):=coeff(expand(product(x+i,i,1,2*(n-1))),x,(n-1)); /* Lorraine Lee, Oct 12 2019 */

PARI

a(n)=polcoeff(prod(k=0,2*n-2,1+k*x),n-1)

PARI

vector(66, n, abs( stirling(2*n-1, n, 1) ) ) /* Joerg Arndt, Jun 09 2012 */

Python

from sympy.functions.combinatorial.numbers import stirling
def A129505(n): return stirling((n<<1)-1,n,kind=1) # Chai Wah Wu, Jun 08 2025

Formula

Unsigned central Stirling numbers of the first kind:

G.f.: A(x) = Sum_{n>=0} a(n)*(2*n-1)!/n!*x^n = B'(x), where B(x) satisfies B(x)^2 = x*log(1/(1-B(x))). - Vladimir Kruchinin, Jun 10 2012

a(n) = ((2*n+1)*(-1)^n*((Sum_{i=1..n-1} (Stirling1(2*i-1,i)*C(2*n,2*i-1)*Stirling1(2*(n-i)+1,n-i))/((n-i)*C(n,i)))-n*Stirling1(2*n-1,n) + Stirling1(2*n,n)))/(n+1). - Vladimir Kruchinin, Feb 28 2013

a(n) ~ (1+2*c)/(8*c*sqrt(Pi*(-1-c))) * (-8*c^2/(exp(1)*(1+2*c)))^n * n^(n-3/2), where c = LambertW(-1,-1/(2*exp(1/2))). - Vaclav Kotesovec, Dec 28 2013

a(n) = abs(C(2*n-1,n-1)*Sum_{i=1..n-1} (Stirling1(n-1,n-i-1)*Stirling1(n,i+1)/C(n-1,i))). - Chai Wah Wu, Jun 08 2025

Extensions

Minor edits by Vaclav Kotesovec, Mar 31 2014